Solution of Derivative

05 300 this implies there is no value to waiting

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Unformatted text preview: we only produce if S > 8 and we would shut down if S < 8; hence S∗ = S ∗ = 8. If there is no cost of undoing options there will be no hysteresis (e.g. producers will never be producing at a loss). Question 17.14. As in Example 17.1, we receive an underlying cash flow worth 300/.03 = 10000 by paying a strike of the mine cost plus the PV of the cost of extraction 250/.05 + 1000 = 6000. Note the dividend implied by the lease rate is 10000 (.03) = 300 is equal to the interest cost on the strike 6000 (.05) = 300. This implies there is no value to waiting. Using the call option method, CallPerpetual (10000, 6000, .0001, ln (1.05) , ln (1.03)) = {4000, 9903.68} . (12) The mine is worth its current NPV. Question 17.16. The value of a producing mine is Vp (S) = S 250 S − 5000 + PutPerpetual , , .2, ln (1.05) , ln (1.03) .03 .03 .05 (13) which is the value of no shut down mine plus the value of being able to shutdown. Working backwards, if S is the trigger price of our mine investment (i.e. we receive Vp S when St = S ), the current value is V S, S = 300 S h1 × Vp S − 1000 (14) Using h1 = 1.5812, we want to choose S to maximize the above quantity. Using Excel’s solver or other numerical program we S = 464.00 and V 300, S = 4830.21. Note the trigger price is lower than when there is no shut down option. 122 Chapter 18 The Lognormal Distribution Question 18.2. If z is standard normal, µ + σ × z is N µ, σ 2 hence our five standard normals can be use to create the desired properties: .8 + 5 (−1.7) = −7.7, .8 + 5 (.55) = 3.55, .8 + 5 (−.3) = −0.7, .8 + 5 (−.02) = .7, and .8 + 5 (.85) = 5.05. Question 18.4. Sums and differences of two random variables are normally distributed hence x1 + x2 is normally distributed with mean µ1 + µ2 = 10 and variance 2 2 σ1 + σ2 + 2ρσ1 σ2 = .5 + 14 + 2 (.5) (14) (−.3) = 10.3 The difference is normally distributed with mean µ1 − µ2 = −6 and (higher) variance 2 2 σ1 + σ2 − 2ρσ1 σ2 = .5 + 14 + 2 (.5) (14) (.3) = 18.7. Question 18.6. If x ˜N µ, σ 2 then E (ex ) = ex µ+.5σ ; using the given numbers, E (ex ) = e2+2.5 = 90.017. There is a 50% probability x is below its mean of 2 hence the median of ex is e2 = 7.3891. 2 Question 18.8. √ Since St = S0 exp α − 1 σ 2 t + σ tz where z is standard normal, 2 P (St > 105) = P =P =P √ 1 105 α − σ 2 t + σ tz > ln 2 100 z> ln −z < 105 100 − ln − α − 1σ2 t 2 √ σt 105 100 + α − 1σ2 t 2 √ σt = N (d2 ) √ 1 where d2 = ln (100/105) + α − 2 σ 2 t / σ t . Using the given parameters, d2 = .045967and N (d2 ) = .4817. For this parameter specification, the probability St > 105 increases with t and 123 Part 5 Advanced Pricing Theory decreases with σ . Analytically, since N (d2 ) > 0, we have the derivative will have the same sign as ∂d2 /∂t and ∂d2 /∂σ . Specifically, α − σ 2 /2 − ln (S0 /K) /t ∂P (St > K) = N (d2 ) >0 √ ∂t 2σ t since α − σ 2 /2 = .035 > 0 and ln (S0 /K) < 0. As an example, if t is 5 years, there is a 57.46% chance of being greater than 105. For volatility, let t = 1. Then ∂P (St > K) α − ln (S0 /K) 1 + = −N (d2 ) 2σ 2 4 ∂σ <0 Question 18.10. We have P (St < 98) = P =P √ 1 98 α − σ 2 t + σ tz < ln 2 100 z< ln 98 100 − α − 1σ2 t 2 √ σt = N (−d2 ) with −d2 = (ln (98/100) − .035) /.3 = −.18401. Hence P (St < 98) = N (−.18401) = 42.70%. Question 18.12. See Figure Two. Option prices depend on the conditional (risk neutral) expectation, not the probability the option is in the money. As T increases, the likelihood that ST > KT may be lower; however, the payoff depends on the conditional expectation (since the option does not pay a constant amount). The increased dispersion offsets the lower probability (for the call option). 124 Chapter 18 The Lognormal Distribution Figure Two (Problem 18.12) 0.9 0.8 0.7 P (ST < 100e.08T) 0.6 0.5 P (ST > 100e.08T) 0.4 0.3 0.2 0 5 10 T 15 20 25 Question 18.14. The mean should be varying year by year; whereas, the standard deviation should be more stable. 125 Chapter 19 Monte Carlo Valuation Question 19.2. The histogram should be similar to a standard normal density (“bell” shaped). Since a uniform 2 distribution has a mean of 0.5 and a variance of 1/12, the mean of 1=1 ui − 6 is zero and the i variance (& standard deviation) will be one since 12 var ui = 12var (ui ) = 1. i =1 Question 19.4. √ The standard deviation of the estimate will be sn / n where sn is the sample standard deviation of the n simulations. Since sn is close to 2.9, n = 84000 should give a standard error close to 0.01. Question 19.6. The simulations should be generated by S1 = 100 exp .06 − .42 /2 + .4z where z is standard normal. The claim prices should be e−.06 S α where α is the relevant power and the S α is the average from the simulations. These values should be close to 100α exp (α − 1) .06 + α2 .4 2 . Using this, the three values (should be close to 12461...
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This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.

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